16-Civ-B7 Transportation Planning and Engineering · December 2013

Question 4 of 7

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 98-Civ-B7 Highway Engineering, National Examinations December 2013 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that a total of five solutions is required, that only the first five as they appear in the answer book will be marked, and that all questions are of equal value. The grading scheme printed on page 1 confirms 20 marks per question, split as: Q1 20; Q2 20; Q3 (a) 15 and (b) 5; Q4 (a) 8 and (b) 12; Q5 (a) 8 and (b) 12; Q6 20; Q7 (a) 8 and (b) 12. All seven printed questions are worked below, because this set is a study resource rather than a timed attempt; on exam day a candidate submits only the first five, in order. The paper also states that any data required but not given may be assumed and that assumptions should be recorded with the answer — several questions need that licence, and every assumption is flagged where it is made.

Reference texts. N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed. (sight distance, vertical and horizontal alignment, traffic stream models, earthwork); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (Canadian design-domain values for stopping sight distance, perception-reaction time and deceleration); AASHTO, A Policy on Geometric Design of Highways and Streets (the tabulated metric stopping sight distances); AASHTO, Guide for Design of Pavement Structures (1993) (rigid pavement thickness, reliability, drainage and load-transfer coefficients); Asphalt Institute, Mix Design Methods MS-2 (gradation charts, the 0.45 power chart, aggregate blending); M.S. Mamlouk and J.P. Zaniewski, Materials for Civil and Construction Engineers, 4th ed. (aggregate moisture states, sieve analysis); Transportation Association of Canada, Pavement Asset Design and Management Guide (Canadian pavement design practice).

Check — assumptions carried through this paper. Four inputs the exam does not supply are assumed under its own Note 2 (“any data required, but not given, can be assumed”), and each is restated at the point of use: (i) Question 2 needs a stopping-sight-distance basis — a 2.5 s perception-reaction time and a 3.4 m/s² deceleration, the TAC and AASHTO design values, giving the tabulated 185 m at 100 km/h; (ii) Question 2 also needs to know whether the 600 m radius is to the road centreline — it is taken as the centreline, and Step 5 shows the alternative reading changes the answer by 0.02 m; (iii) Question 6 does not say whether the transverse joints are dowelled — dowels are assumed, giving a load-transfer coefficient J = 3.2, with the undowelled case quantified in a callout; (iv) Question 6 gives a drainage description rather than a coefficient, so C_d = 1.00 is read from the AASHTO table, again with the alternative quantified.

Question 4 20 marks — (a) 8, (b) 12

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Cross-sectional cut and fill areas at five stations spaced 1000 m apart, with a grade point at the middle station where both areas vanish:

Given data — cross-sectional areas
Station (m)	Distance from 840+000 (m)	Cut area (m²)	Fill area (m²)
840+000.000	0	80	—
841+000.000	1000	70	—
842+000.000	2000	0	0
843+000.000	3000	—	60
844+000.000	4000	—	70

Find. The total volume of cut and the total volume of fill between stations 840+000.000 and 844+000.000, computed first by the average end area method and then by the prismoidal formula.

Cut and fill areas along the alignment, with the grade point at station 842+000 separating the two solids.

Approach. Split the alignment at the grade point at station 842+000, where the section changes from all cut to all fill, and integrate each branch separately — by the trapezoidal rule for the average end area method, and by Simpson’s rule (which is what the prismoidal formula is) over the pair of intervals in each branch.

Establish the interval length and locate the grade point. The stations are printed 1000.000 m apart, so each interval is $L = 1000$ m. At station 842+000 both the cut area and the fill area are zero: this is the grade point, where the finished profile crosses the existing ground. Cut therefore occupies the first two intervals only and fill the last two only, and the two must never be averaged across the grade point.
Apply the average end area method to the cut branch. The method treats the solid between two sections as a prism whose area is the mean of the end areas, $V = L(A_1 + A_2)/2$: $$V_{840-841} = 1000\left(\frac{80 + 70}{2}\right) = 75\,000\text{ m}^3$$ $$V_{841-842} = 1000\left(\frac{70 + 0}{2}\right) = 35\,000\text{ m}^3$$ $$V_{cut,\,AEA} = 75\,000 + 35\,000 = \boxed{110\,000\text{ m}^3}$$
Apply the same method to the fill branch. $$V_{842-843} = 1000\left(\frac{0 + 60}{2}\right) = 30\,000\text{ m}^3, \qquad V_{843-844} = 1000\left(\frac{60 + 70}{2}\right) = 65\,000\text{ m}^3$$ $$V_{fill,\,AEA} = 30\,000 + 65\,000 = \boxed{95\,000\text{ m}^3}$$
Set up the prismoidal formula correctly. The prismoidal formula $$V = \frac{L'}{6}\left(A_1 + 4A_m + A_2\right)$$ applies to a solid of total length $L'$ with a <em>middle</em> section $A_m$ measured midway between the ends, so it consumes sections in pairs of intervals and needs an odd number of sections. Each branch here has exactly three sections, so each branch forms a single prismoid of length $L' = 2 \times 1000 = 2000$ m with the intermediate station supplying $A_m$.
Evaluate the prismoidal volume for the cut branch. With $A_1 = 80$ m<sup>2</sup> at 840+000, $A_m = 70$ m² at 841+000 and $A_2 = 0$ at the grade point, $$V_{cut,\,prism} = \frac{2000}{6}\bigl[80 + 4(70) + 0\bigr] = 333.333(360) = \boxed{120\,000\text{ m}^3}$$
Evaluate the prismoidal volume for the fill branch. With $A_1 = 0$ at the grade point, $A_m = 60$ m² at 843+000 and $A_2 = 70$ m² at 844+000, $$V_{fill,\,prism} = \frac{2000}{6}\bigl[0 + 4(60) + 70\bigr] = 333.333(310) = \boxed{103\,333\text{ m}^3}$$
Compare the two methods and interpret the difference. The prismoidal result exceeds the end-area result by 9.1% in cut and 8.8% in fill. That direction is worth noting, because the textbook shorthand “end areas overestimate” is only true when the intermediate section is smaller than the average of the two ends. Here the opposite holds: on the cut branch the mid-station area of 70 m² is far larger than the mean of 80 and 0, so a straight-line interpolation between the end sections understates the solid, and the prismoidal formula — which fits a parabola through the three areas — correctly picks up the bulge.
State the earthwork balance. Subtracting fill from cut, $$\text{Surplus} = 110\,000 - 95\,000 = 15\,000\text{ m}^3 \text{ (end area)}, \qquad 120\,000 - 103\,333 = 16\,667\text{ m}^3 \text{ (prismoidal)}$$ so the 4 km section generates a surplus of roughly 15 000 to 17 000 m³ of material, before any allowance for shrinkage on compaction or swell of rock. Applied at a typical 10% shrinkage for common earth, the surplus available for disposal or for adjacent fill would be closer to 4000 to 8000 m³.

Check — the section spacing is as printed. The exam prints the five stations 1000.000 m apart, which is far wider than the 20 m to 50 m normally used for earthwork quantities. The volumes above are computed on the data as given. In practice, sections a kilometre apart cannot resolve real ground, and the two methods differing by 9% is itself the symptom — with 20 m sections the two would agree to well under 1%. Quote the numbers, but state that the estimate is a preliminary quantity, not a pay quantity.

Final results — Question 4
Quantity	(a) Average end area	(b) Prismoidal formula
Cut, 840+000 to 842+000	110 000 m³	120 000 m³
Fill, 842+000 to 844+000	95 000 m³	103 333 m³
Net surplus of cut over fill	15 000 m³	16 667 m³
Difference from the end-area value	—	+9.1% cut, +8.8% fill

Topic. Earthwork quantities from cross-sectional areas: the average end area method, the prismoidal formula, and the grade point that separates cut from fill.

Key relations. Average end area, $V = L(A_1 + A_2)/2$; prismoidal, $V = (L'/6)(A_1 + 4A_m + A_2)$ over a length $L'$ spanning two intervals; prismoidal correction for a level-section solid, $V_{corr} = (L/6)(A_1 - 2A_m + A_2)$; and the mass-haul surplus, cut minus fill, adjusted by a shrinkage factor before it can be reused as embankment.

Why this works. Both formulae are numerical integrations of the cross-sectional area along the alignment. The end area method is the trapezoidal rule, exact only when the area varies linearly with chainage. The prismoidal formula is Simpson’s rule, exact whenever the area varies as a quadratic in chainage — which is precisely the case for a prismatoid, a solid whose parallel end faces are joined by plane surfaces. Since a road cross-section formed by constant side slopes on a uniformly changing ground line does vary quadratically, the prismoidal result is the theoretically correct one and the end-area value is the approximation.

Common pitfall. The dominant error is averaging a cut area against a fill area across the grade point, which manufactures volume that does not exist; the zero-area station must be used as a hard boundary. The second is applying the prismoidal formula to an even number of sections, or using it interval by interval with $A_m$ taken as the mean of the ends — that substitution collapses it back into the end-area formula and gains nothing. The third is forgetting that cut and fill volumes are not directly comparable until shrinkage or swell is applied.

Source. N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed., Chapter 3, cross sections and earthwork volumes; Transportation Association of Canada, Geometric Design Guide for Canadian Roads, cross-section elements and grading.

← Question 3 Next: Question 5 →